leetcode 19. Remove Nth Node From End of List 删除倒数第n个节点；

Given a linked list, remove the n-th node from the end of list and return its head.

Example:

Given linked list: 1->2->3->4->5, and n = 2.

After removing the second node from the end, the linked list becomes 1->2->3->5.

Note:

Given n will always be valid.

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* removeNthFromEnd(ListNode* head, int n) 
    {
        if(!head)
            return nullptr;
        if(n==0)
            return head;
        ListNode *dummy=new ListNode (0);
        dummy->next=head;
        ListNode *fast=dummy;
        ListNode *slow=dummy;
        for(int i=0;i<n;i++)
            fast=fast->next;
        
        while(fast->next)
        {
            fast=fast->next;
            slow=slow->next;
        }
        slow->next=slow->next->next;
        return dummy->next;
        
    }
};

思路：双指针 fast slow；

fast先走n步而slow不变；都保持在dummy处；

然后fast和slow一起走 判断fast->next!=nullptr;

CODE: